Question

The pressure exerted by $6.0 \, g$ of methane gas in a $0.03\, m^3$ vessel at $129^{\circ}C$ is (Atomic masses : $C = 12.01, H = 1.01$ and $R = 8.314 \, JK^{-1} mol^{-1}$)

• A. 13409 Pa
• B. 41648 Pa
• C. 31684 Pa
• D. 215216 Pa

Answer 1

Answer: (B)

41648 Pa

Answer 2

Given, volume, $V=0.03\,\,{{m}^{3}}$ temperature,
$T = 129 + 273 = 402$
$K$ mass of methane,
$w = 6.0\, g\, mol.$ mass of methane,
$M=12.01+4\times 1.01=16.05$
From, ideal gas equation,
$pV=nRT$
$p=\frac{6}{16.05}\times \frac{8.314\times 402}{0.03} = 41648Pa$